y^'+20xy=0

Lösung

y^{^{'}} + 20 x y = 0

y = e^{- 10 x^{2} + c_{1}}

Schritte zur Lösung

y^{'} (x) + 20 x y = 0

Solve separable ODE:

y = e^{- 10 x^{2} + c_{1}}

y = e^{- 10 x^{2} + c_{1}}

(2 y + y x^{2}) d y = (x + x y^{2}) d x

y^{^{'}} = 4 e^{3 x} e^{- 2 y}

5 x + y y^{^{'}} = 0

\frac{( x ^{2} + 4 ) d y}{d x} + 3 x y = x

(x - 2) y^{^{'}} + y = 0